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Hatshy [7]
3 years ago
7

Two circuits are created using two identical light bulbs. In the circuit A, the bulbs are hooked up in series. In circuit B, the

light bulbs are hooked up parallel. Each circuit is powered by a 6-volt battery.
Which statement or observation would not be supported by the two circuits?
A) The current in the circuit A would be less.
B) The light bulbs in the circuit B would be brighter.
C) The light bulbs in circuit B are equal in brightness.
D) If one light burns out in circuit A, the other bulbs gets brighter.
Computers and Technology
2 answers:
Ulleksa [173]3 years ago
5 0
C is not correct since in the case of circuits that are lined up in series, if one or them dies they all die.
kirill115 [55]3 years ago
5 0

Answer:

D) If one light burns out in the circuit A, the other bulb gets brighter.

Explanation:

USATestprep

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Um?<br><br> i went to check my questions and i found this-
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just go ahead and refresh the page

Explanation:

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3 years ago
SHOW ALL YOUR WORK. REMEMBER THAT PROGRAM SEGMENTS ARE TO BE WRITTEN IN JAVA. Assume that the classes listed in the Java Quick R
Ne4ueva [31]

Answer:

Check the explanation

Explanation:

CODE:-

import java.util.*;

class UserName{

  ArrayList<String> possibleNames;

  UserName(String firstName, String lastName){

      if(this.isValidName(firstName) && this.isValidName(lastName)){

          possibleNames = new ArrayList<String>();

          for(int i=1;i<firstName.length()+1;i++){

              possibleNames.add(lastName+firstName.substring(0,i));

          }  

      }else{

          System.out.println("firstName and lastName must contain letters only.");

      }

  }

  public boolean isUsed(String name, String[] arr){

      for(int i=0;i<arr.length;i++){

          if(name.equals(arr[i]))

              return true;

      }

      return false;

  }

  public void setAvailableUserNames(String[] usedNames){

      String[] names = new String[this.possibleNames.size()];

      names = this.possibleNames.toArray(names);

      for(int i=0;i<usedNames.length;i++){

          if(isUsed(usedNames[i],names)){

              int index = this.possibleNames.indexOf(usedNames[i]);

              this.possibleNames.remove(index);

              names = new String[this.possibleNames.size()];

              names = this.possibleNames.toArray(names);

          }

      }

  }

  public boolean isValidName(String str){

      if(str.length()==0) return false;

      for(int i=0;i<str.length();i++){

          if(str.charAt(i)<'a'||str.charAt(i)>'z' && (str.charAt(i)<'A' || str.charAt(i)>'Z'))

              return false;

      }

      return true;

  }

  public static void main(String[] args) {

      UserName person1 = new UserName("john","smith");

      System.out.println(person1.possibleNames);

      String[] used = {"harta","hartm","harty"};

      UserName person2 = new UserName("mary","hart");

      System.out.println("possibleNames before removing: "+person2.possibleNames);

      person2.setAvailableUserNames(used);

      System.out.println("possibleNames after removing: "+person2.possibleNames);

  }

}

Kindly check the attached image below for the code output.

5 0
4 years ago
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olya-2409 [2.1K]

Answer:

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Explanation:

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The login details and some important credentials to access user data contains both the user's public key data and private key data. Both private key and public key are two keys that work together to accomplish security goals.

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Answer:

boolean isEven = false;

if (x.length % 2 == 0)

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Comparable currentMax;

int currentMaxIndex;

for (int i = x.length - 1; i >= 1; i--)

{

currentMax = x[i];

currentMaxIndex = i;

for (int j = i - 1; j >= 0; j--)

{

if (((Comparable)currentMax).compareTo(x[j]) < 0)

{

currentMax = x[j];

currentMaxIndex = j;

}

}

x[currentMaxIndex] = x[i];

x[i] = currentMax;

}

Comparable a = null;

Comparable b = null;

if (isEven == true)

{

a = x[x.length/2];

b = x[(x.length/2) - 1];

if ((a).compareTo(b) > 0)

m = a;

else

m = b;

}

else

m = x[x.length/2];

8 0
4 years ago
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